What is the integral of 1/(1000-x) ?

The integral of 1/(1000-x) is -ln|1000-x|+C

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integral of 1/(1000-x)

\int \frac{1}{1000 - x} d x

- ln ∣ 1000 - x ∣ + C

Solution steps

\int \frac{1}{1000 - x} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1000 - x

= - ln ∣ 1000 - x ∣

Add a constant to the solution

= - ln ∣ 1000 - x ∣ + C

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\frac{d}{d x} (x^{3} - 2 x^{2} + 2)

\frac{\partial}{\partial y} (x cos (y) sin (z))

2 y^{^{'}} + 10 y = 80, y (0) = 0

What is the integral of 1/(1000-x) ?
The integral of 1/(1000-x) is -ln|1000-x|+C